|
* [[밀스 상수]] 또한 무리수로 입증되지 않았다.
* [[코플랜드 에르되시 상수]]는 소수점 표기를 연결하여 형성된다.
==Sketch of a proof that {{mvar|e}} is transcendental==
The first proof that [[E (mathematical constant)|the base of the natural logarithms, {{mvar|e}}]], is transcendental dates from 1873. We will now follow the strategy of [[David Hilbert]] (1862–1943) who gave a simplification of the original proof of [[Charles Hermite]]. The idea is the following:
Assume, for purpose of finding a contradiction, that {{mvar|e}} is algebraic. Then there exists a finite set of integer coefficients ''c''<sub>0</sub>, ''c''<sub>1</sub>, ..., ''c<sub>n</sub>'' satisfying the equation:
:<math>c_{0}+c_{1}e+c_{2}e^{2}+\cdots+c_{n}e^{n}=0, \qquad c_0, c_n \neq 0.</math>
Now for a positive integer ''k'', we define the following polynomial:
:<math> f_k(x) = x^{k} \left [(x-1)\cdots(x-n) \right ]^{k+1},</math>
and multiply both sides of the above equation by
:<math>\int^{\infty}_{0} f_k e^{-x}\,dx,</math>
to arrive at the equation:
:<math>c_{0} \left (\int^{\infty}_{0} f_k e^{-x}\,dx\right )+ c_1e\left ( \int^{\infty}_{0}f_k e^{-x}\,dx\right )+\cdots+ c_{n}e^{n} \left (\int^{\infty}_{0}f_k e^{-x}\,dx\right ) = 0.</math>
By splitting respective domains of integration, this equation can be written in the form
:<math>P+Q=0</math>
where
:<math>\begin{align}
P &= c_{0}\left ( \int^{\infty}_{0}f_k e^{-x}\,dx\right )+ c_{1}e\left (\int^{\infty}_{1}f_k e^{-x}\,dx\right )+ c_{2}e^{2}\left (\int^{\infty}_{2}f_k e^{-x}\,dx\right ) +\cdots+ c_{n}e^{n}\left (\int^{\infty}_{n}f_k e^{-x}\,dx\right ) \\
Q &= c_{1}e\left (\int^{1}_{0} f_k e^{-x}\,dx\right )+c_{2}e^{2} \left (\int^{2}_{0} f_k e^{-x}\,dx\right )+\cdots+c_{n}e^{n}\left (\int^{n}_{0} f_k e^{-x}\,dx \right )
\end{align}</math>
'''Lemma 1.''' For an appropriate choice of ''k'', <math>\tfrac{P}{k!}</math> is a non-zero integer.
<blockquote>'''Proof.''' Each term in ''P'' is an integer times a sum of factorials, which results from the relation
:<math>\int^{\infty}_{0}x^{j}e^{-x}\,dx=j!</math>
which is valid for any positive integer ''j'' (consider the [[Gamma function]]).
It is non-zero because for every ''a'' satisfying 0< ''a'' ≤ ''n'', the integrand in
:<math>c_{a}e^{a}\int^{\infty}_{a} f_k e^{-x}\,dx</math>
is ''e<sup>−x</sup>'' times a sum of terms whose lowest power of ''x'' is ''k''+1 after substituting ''x'' for ''x''+''a'' in the integral. Then this becomes a sum of integrals of the form
:<math>A_{j-k}\int^{\infty}_{0}x^{j}e^{-x}\,dx</math> Where ''A''<sub>j-k</sub> is integer.
with ''k''+1 ≤ ''j'', and it is therefore an integer divisible by (''k''+1)!. After dividing by ''k!'', we get zero [[Modular arithmetic|modulo]] (''k''+1). However, we can write:
:<math>\int^{\infty}_{0} f_k e^{-x}\,dx = \int^{\infty}_{0} \left ( \left [(-1)^{n}(n!) \right ]^{k+1}e^{-x}x^k + \cdots \right ) dx</math>
and thus
:<math>{\frac {1}{k!}}c_{0}\int _{0}^{\infty }f_{k}e^{-x}\,dx\equiv c_{0}[(-1)^{n}(n!)]^{k+1}\not\equiv 0{\pmod {k+1}}.</math>
So when dividing each integral in ''P'' by ''k!'', the initial one is not divisible by ''k''+1, but all the others are, as long as ''k''+1 is prime and larger than ''n'' and |''c''<sub>0</sub>|. It follows that <math>\tfrac{P}{k!}</math> itself is not divisible by the prime ''k''+1 and therefore cannot be zero.</blockquote>
'''Lemma 2.''' <math>\left|\tfrac{Q}{k!}\right|<1</math> for sufficiently large <math>k</math>.
<blockquote> '''Proof.''' Note that
:<math>\begin{align}
f_k e^{-x} &= x^{k}[(x-1)(x-2)\cdots(x-n)]^{k+1}e^{-x}\\
&= \left (x(x-1)\cdots(x-n) \right)^k \cdot \left ((x-1)\cdots(x-n)e^{-x}\right)\\
&= u(x)^k \cdot v(x)
\end{align}</math>
where <math>u(x)</math> and <math>v(x)</math> are continuous functions of <math>x</math> for all <math>x</math>, so are bounded on the interval <math>[0,n]</math>. That is, there are constants <math>G, H > 0</math> such that
:<math> \left |f_k e^{-x} \right | \leq |u(x)|^k \cdot |v(x)| < G^k H \quad \text{ for } 0 \leq x \leq n.</math>
So each of those integrals composing <math>Q</math> is bounded, the worst case being
:<math>\left|\int_{0}^{n}f_{k}e^{-x}\,dx\right| \leq \int_{0}^{n} \left |f_{k}e^{-x} \right |\,dx \leq \int_{0}^{n}G^k H\,dx = nG^k H.</math>
It is now possible to bound the sum <math>Q</math> as well:
:<math>|Q| < G^{k} \cdot nH \left (|c_1|e+|c_2|e^2+\cdots+|c_n|e^{n} \right ) = G^k \cdot M,</math>
where <math>M</math> is a constant not depending on <math>k</math>. It follows that
:<math>\left| \frac{Q}{k!} \right| < M \cdot \frac{G^k}{k!} \to 0 \quad \text{ as } k \to \infty,</math>
finishing the proof of this lemma.</blockquote>
Choosing a value of <math>k</math> satisfying both lemmas leads to a non-zero integer (<math>P/k!</math>) added to a vanishingly small quantity (<math>Q/k!</math>) being equal to zero, is an impossibility. It follows that the original assumption, that {{mvar|e}} can satisfy a polynomial equation with integer coefficients, is also impossible; that is, {{mvar|e}} is transcendental.
===The transcendence of {{pi}}===
A similar strategy, different from [[Ferdinand von Lindemann|Lindemann]]'s original approach, can be used to show that the [[Pi|number {{pi}}]] is transcendental. Besides the [[gamma-function]] and some estimates as in the proof for {{mvar|e}}, facts about [[symmetric polynomial]]s play a vital role in the proof.
For detailed information concerning the proofs of the transcendence of {{pi}} and {{mvar|e}}, see the references and external links.
== 같이 보기 ==
| 